Geo4.110

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Find the circle which passes through the points of intersection of the circles x^{2}+y^{2}+6x+4y-12=0,x^{2}+y^{2}-4x-6y-12=0\, and cuts the circle x^{2}+y^{2}-2x+3=0\, orthogonally.

Equation of common chord is

(x^{2}+y^{2}+6x+4y-12)-(x^{2}+y^{2}-4x-6y-12)=0\,

10x+10y=0,x+y=0\,

Equation of the circle passing through the points of intersection of the circles is

x^{2}+y^{2}+6x+4y-12+k(x+y)=0\,

x^{2}+y^{2}+x(6+k)+y(4+k)-12=0\,

This circle cuts the third given circle orthogonally, hence

2(-1)({\frac  {6+k}{2}})+2(0)({\frac  {4+k}{2}})=-12+3\,

-6-k=-9,k=3\,

Therefore the required circle is

x^{2}+y^{2}+9x+7y-12=0\,

Main Page:Geometry:Circles